Tangent at the point P 1 ≡ (a , a 3 +1), (a ≠ 0) on the curve y= x 3 +1 meets the curve again at P 2 . The tangent at P 2 meets the curve again at P 3 and so on.
Find 
& 
, where x(P i ) & y(P i ) are the abscissa and ordinate of P i respectively.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
Sol. We have,
y = x 3 + 1
⇒ 
= 3x 2 ⇒ 
= 3a
2 ,
The equation of the tangent at P 1 (a, a 3 + 1) is
y – (a 3 + 1) = 3a 2 (x – a) ... (i)
This meets the curve at P 2 . So, let the coordinates of P 2 be (a 2 , a 2 3 + 1)
As P 2 lies on (i), therefore
a 2 3 + 1 –(a 3 + 1) = 3a 2 (a 2 –a)
⇒ a 2 3 –a 3 = 3a 2 (a 2 –a)
⇒ (a 2 –a) 2 (a 2 + 2a) = 0
⇒ a 2 = –2a [ a 2 ≠ a]
So, the coordinates of P 2 are (–2a, –8a 3 + 1)
Comparing this with the coordinates of P 1 , we find that the coordinates of P 2 can be obtained from the coordinates of P 1 if we replace a by –2a.
Similarly, coordinates of P 3 , P 4 ,..... are
P 3 (4a, (4a) 3 + 1), P 4 (–8a, (–8a) 3 + 1) and so on.
Now,
= a –2a + 4a –8a + ...... upto 2n terms
= a 
= 
(1 –4 n )
and,
= (a 3 + 1) + (–8a 3 + 1 + (64a 3 + 1) + .......2n terms
= a 3 (1 –8 + 64 + .....) + 2n
= a 3 
+ 2n
= 
(1 –64 n ) + 2n
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